integral of x^5e^{3x}

解答

\int x^{5} e^{3 x} d x

\frac{1}{729} (243 e^{3 x} x^{5} - 5 (81 e^{3 x} x^{4} - 4 (27 e^{3 x} x^{3} - 3 (9 e^{3 x} x^{2} - 2 (3 e^{3 x} x - e^{3 x}))))) + C

求解步骤

\int x^{5} e^{3 x} d x

使用换元积分法

= \int \frac{e ^{u} u ^{5}}{729} d u

提出常数:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{729} \cdot \int e^{u} u^{5} d u

使用分布积分法

= \frac{1}{729} (u^{5} e^{u} - \int 5 u^{4} e^{u} d u)

\int 5 u^{4} e^{u} d u = 5 (u^{4} e^{u} - 4 (u^{3} e^{u} - 3 (u^{2} e^{u} - 2 (e^{u} u - e^{u}))))

= \frac{1}{729} (u^{5} e^{u} - 5 (u^{4} e^{u} - 4 (u^{3} e^{u} - 3 (u^{2} e^{u} - 2 (e^{u} u - e^{u})))))

u = 3 x

代回

= \frac{1}{729} ((3 x)^{5} e^{3 x} - 5 ((3 x)^{4} e^{3 x} - 4 ((3 x)^{3} e^{3 x} - 3 ((3 x)^{2} e^{3 x} - 2 (e^{3 x} \cdot 3 x - e^{3 x})))))

化简

\frac{1}{729} ((3 x)^{5} e^{3 x} - 5 ((3 x)^{4} e^{3 x} - 4 ((3 x)^{3} e^{3 x} - 3 ((3 x)^{2} e^{3 x} - 2 (e^{3 x} \cdot 3 x - e^{3 x}))))) : \frac{1}{729} (243 e^{3 x} x^{5} - 5 (81 e^{3 x} x^{4} - 4 (27 e^{3 x} x^{3} - 3 (9 e^{3 x} x^{2} - 2 (3 e^{3 x} x - e^{3 x})))))

= \frac{1}{729} (243 e^{3 x} x^{5} - 5 (81 e^{3 x} x^{4} - 4 (27 e^{3 x} x^{3} - 3 (9 e^{3 x} x^{2} - 2 (3 e^{3 x} x - e^{3 x})))))

解答补常数

= \frac{1}{729} (243 e^{3 x} x^{5} - 5 (81 e^{3 x} x^{4} - 4 (27 e^{3 x} x^{3} - 3 (9 e^{3 x} x^{2} - 2 (3 e^{3 x} x - e^{3 x}))))) + C

d er i v a t i v e f (x) = (x + 3)^{2}

\int_{\frac{1}{2}}^{1} (x^{- 3} - 2) d x

t an g e n t y = 6 x e^{x}, (0, 0)

d er i v a t i v e cos (4 t)

\frac{\partial}{\partial x} (e^{- y} (x + e^{y}))